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Let $D$ be an integral domain and let $I$ be an ideal in $D$ not containing a non-unit element $x$. Is it true that there is a prime ideal $P$, containing $I$, such that $x \not \in P$? What about a maximal ideal?

This is true (for the prime case) if $I=(0)$, as has been shown in a question posed on this site, but the argument uses Zorn’s Lemma in a way which can’t be directly adapted to this scenario.

In case the answer is negative, is there any condition we can impose on $D$ so that it is true? Maybe a PID? I couldn’t quite figure out what might be necessary…

Thanks in advance!

Gauss
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  • The maximal ideal question is certainly not true if $D$ is a local domain, since it will have only one maximal ideal which must contain $x$. Furthermore, if $I$ is prime, then you can use the argument for $I=(0)$ in $D/I$ to obtain a prime $P$ of $D$ such that $P+I$ is prime and does not contain $x+I$, then $P$ will not contain $x$ in $D$. – walkar Oct 13 '22 at 13:14
  • Another thought -- if $x$ is not in the radical $\sqrt{I}$, then there must be a minimal prime ideal of $\sqrt{I}$ which does not contain $x$. If $x$ is in $\sqrt{I}$, then every minimal prime over $I$ contains $x$, and you might be in trouble. Check this answer: https://math.stackexchange.com/a/3959804/98077 – walkar Oct 13 '22 at 13:20
  • @walkar Thanks for the thoughts and for the reference! – Gauss Oct 13 '22 at 17:19

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